prove $a,b,c$ in A.P if $\tan\dfrac{A}{2}=\dfrac{5}{6}$ and $\tan\dfrac{C}{2}=\dfrac{2}{5}$

Question

In $\Delta ABC$, if $\tan\dfrac{A}{2}=\dfrac{5}{6}$ and $\tan\dfrac{C}{2}=\dfrac{2}{5}$, then prove that the sides $a,b,c$ are in A.P.

My Attempt $$ \sin A=\frac{2.5}{6}.\frac{36}{61}=\frac{60}{61}\\ \sin C=\frac{2.2}{5}.\frac{25}{29}=\frac{20}{29}\\ $$ it is solved in my reference some fomula involving $2s=a+b+c$, can I prove it using the basic known properties of triangles ?

Answer 1

$$\dfrac1{\tan\dfrac B2}=\cot B/2=\tan(A/2+C/2)=?$$

Use

$$\sin2x=\dfrac{2\tan x}{1+\tan^2x}=?$$ for $2x=A,B,C$

Answer 2

It is $$\tan(\alpha/2)=\frac{r}{s-a}$$ and $$\tan(\gamma/2)=\frac{r}{s-c}$$ where $$s=\frac{a+b+c}{2}$$ so we get $$\frac{5}{6}(s-a)=\frac{2}{5}(s-c)$$ Can you finish now?

Answer 3

Hint:

Like In $\Delta ABC$, find $\cot\dfrac{B}{2}.\cot\dfrac{C}{2}$ if $b+c=3a$,

$$2b=a+c$$ will hold true if $\tan\dfrac A2\tan\dfrac B2=\dfrac13$

Answer 4

You can first deduce $$ \tan\frac{B}{2}=\tan\left(\frac{\pi}{2}-\frac{A+C}{2}\right)=\cot\frac{A+C}{2}= \frac{1-\tan\frac{A}{2}\tan\frac{C}{2}}{\tan\frac{A}{2}+\tan\frac{C}{2}}=\frac{2/3}{37/30} =\frac{20}{37} $$ Therefore $$ \sin A=\frac{2(5/6)}{1+25/36}=\frac{60}{61} $$ Similarly, $$ \sin B=\frac{1480}{1769}\qquad \sin C=\frac{20}{29} $$ By the sine law, $$ \frac{a+c}{2}=\frac{b}{2\sin B}(\sin A+\sin C)= b\frac{1769}{2960}\left(\frac{60}{61}+\frac{20}{29}\right)=b $$